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\title{Caculation of SFR}

\author[1,2]{Haotian Song}
\author[1,$\dagger$]{Zhaoyu Zuo}
\affil[1]{Physics Department, Xi'an Jiaotong University, Xi'an, Shaanxi 710049, China}
\affil[2]{School of Physics \& Astronomy, University of Manchester, Manchester M13 9PL, United Kingdom}
\affil[$\dagger$]{taopeng@tamu.edu}
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\begin{abstract}
This PDF show the process of caculating SFR and $\delta N$
 
\end{abstract}
\begin{document}
\maketitle
\section{SFR}

I am not sure whether SFR includes all stars or binaries only, so I caculate it individually.

$S_b$and $S_s$ are in units of Num/yr.

SFR is in units of $M_\odot/yr$

The main idea of caculation is as following.
\begin{equation}
    \label{eq:1}
    S_b \dot (average of M_{low} and M_{up}) + S_b \dot (average M_2) + S_s \dot(average mass) = SFR
\end{equation}

Equation\ref{eq:1}can be written as following, too.

\begin{equation}
    S_b \dot \int_{M_{low}}^{M_{up}} m_1 \epsilon({m_1})dm_1 + S_b \int_{M_{low}}^{M_{up}} \epsilon({m_1}) (\int_{M_{low}}^{M_1}\frac{m_2}{m_1}  dm_2) dm_1 + S_s \int_{M_{low}}^{M_{up}} m \epsilon({m})dm = SFR
\end{equation}


\begin{equation}
    S_b \dot \int_{M_{low}}^{M_{up}} m_1 \epsilon({m_1})dm_1 + S_b \int_{M_{low}}^{M_{up}} \epsilon({m_1}) m_1 \overline{q}  dm_1 + S_s \int_{M_{low}}^{M_{up}} m \epsilon({m})dm = SFR
\end{equation}

chosen q in my caculation is 0.08, 0.18, 0.28... 0.98, so $\overline{q}=0.535$

If I chose $M_{low}=5M_\odot$, as\cite{grimm2003high} .
\begin{equation}
    SFR=(S_b(1+\overline{q}) + S_s)\int_{M_{low}}^{\infty} m \epsilon(m)dm
\end{equation}

However, the Initial Mass Function differs.
When $m>1M_\odot$, $\epsilon(m)=am^{-\alpha}$

For \cite{kroupa2001variation}, $\alpha=+2.3\pm0.7$

For \cite{kroupa1993distribution} $\alpha=+2.7$

\begin{equation}
    SFR=(S_b (1+\overline{q}) + S_s)\int_{M_{low}}^{\infty} am^{1-\alpha}dm
\end{equation}
\begin{equation}
    SFR=(S_b (1+\overline{q}) + S_s) a\frac{m^{2-\alpha}}{2-\alpha}\bigg|_{M_{low}}^{\infty}
\end{equation}
\begin{equation}
    SFR=(S_b (1+\overline{q}) + S_s) a\frac{M_{low}^{2-\alpha}}{\alpha-2}
\end{equation}

For my BPS caculation, $M_{low}=2M_\odot$, so
\begin{equation}
\overline{S_b}=S_b\frac{\int_2^\infty\epsilon(m)dm}{\int_5^\infty\epsilon(m)dm}=S_b(\frac{2}{5})^{1-\alpha}
\end{equation}

And the relation between $S_b$ and $S_s$ is 
$\frac{2S_b}{S_s+2S_b}=f=0.5$
where f is the fraction of binaries and all stars. If $f=0.5$, then $S_s=2S_b$

And we will get a in the following.

As is shown before,  $\epsilon(m)=am^{-\alpha}$. And  $\epsilon(m)$ should be normalized from $M_{low}$ to $M_{max}$, so a is changed and not equal to the Initial a in the essay.
$$1=\int_{M_{low}}^{M_{max}}am^{-\alpha}dm$$

$$1=a\frac{m^{1-\alpha}}{1-\alpha}\bigg|_{M_{low}}^{\infty}=a\frac{M_{low}^{1-\alpha}}{1-\alpha}$$

$$a=\frac{\alpha-1}{M_{low}^{1-\alpha}}$$

In conclution, $S_b$ can be written as following.
\begin{equation}
    \overline{S_b}=(\frac{2}{5})^{1-\alpha}\frac{(\alpha-2)SFR}{(\overline{3}+\overline{q})M_{low}(\alpha-1)}
\end{equation}
where $\overline{3}$ means whether SFR contains single star. If not, $\overline{3}=1$.

\begin{table}[h!]
\centering
\begin{tabular}{lll}
         & $\alpha=2.3$ & $\alpha=2.7$ \\
Binary   & 4.1560       & 10.6984      \\
All star & 1.8046       & 4.6455
      
\end{tabular}
\end{table}
%\noindent\textbf{Disclosures.} The authors declare no conflicts of interest.
\section{Caculate $\delta N$ }
According to \cite{hurley2002evolution}, the $\delta r$ can be written as:
\begin{equation}
\delta r_{j}=S \Phi\left(\ln M_{1 j}\right) \varphi\left(\ln M_{2 j}\right) \Psi\left(\ln a_{j}\right) \delta \ln M_{1} \delta \ln M_{2} \delta \ln a
\end{equation}
and $\delta N = \delta r \delta T$, so
\begin{equation}
    \delta N = \overline{S_b} M_1 \epsilon({m_1}) \delta \ln M_1 k_q \delta q \dot k_a \delta \ln a \delta T
\end{equation}
And the q and ln(a) are mean distribution, so $k_q\delta q = 1/(q_{number})$,$k_a\delta \ln a=1/a_{number}$
where $q_{number}$ and $a_{number}$ is the cycle index in my caculation. $1/a_{number}$ is the length of step after normalization.
\begin{equation}
    \delta N = \overline{S_b} a M_1^{1-\alpha} \delta \ln M_1 k_q \delta q \dot k_a \delta \ln a \delta T
\end{equation}

\begin{equation}
    \delta N = \overline{S_b} (\alpha-1) (\frac{M_1}{2})^{1-\alpha} \frac{\ln(M_{max})-\ln(M_{min})}{M_{number}} \frac{1}{q_{number}} \frac{1}{a_{number}} \delta T
\end{equation}


\section{Adding A to the formula}

The sum of all Success cases should be 1, 
\begin{equation}
    \Sigma Success  = 1,\iiint M_1 \epsilon({m_1}) \delta \ln M_1 k_q \delta q \dot k_a \delta \ln a=1
\end{equation}
but it is not because of failure cases. So A is added.
\begin{equation}
    \Sigma Success + \Sigma Failure  = A \iiint M_1 \epsilon({m_1}) \delta \ln M_1 k_q \delta q \dot k_a \delta \ln a 
\end{equation}
\begin{equation}
    \Sigma Success + \Sigma Failure  = 1 + \Sigma Failure
\end{equation}
\begin{eqnarray}
    A=\frac{1}{ \iiint - \Sigma Failure}=A
\end{eqnarray}

\begin{equation}
    \delta N =(1+\Sigma Failure) \overline{S_b} (\alpha-1) (\frac{M_1}{2})^{1-\alpha} \frac{\ln(M_{max})-\ln(M_{min})}{M_{number}} \frac{1}{q_{number}} \frac{1}{a_{number}} \delta T
\end{equation}

And I got $A=1.0295$
\bibliography{sample}


\end{document}